A body is placed on a smooth inclined plane of inclination 1 in x. The horizontal acceleration to be given to the inclined plane so that the body on it remains at rest with respect to inclined plane is

Options

A. `gsqrt(x^(2)-1)`
B. `(gsqrt(x^(2)-1))/x`
C. `(gx)/sqrt(x^(2)-1)`
D. `g/sqrt(x^(2)-1)`

Answer

To solve the problem step by step, we need to analyze the forces acting on the body placed on the inclined plane and determine the required horizontal acceleration of the inclined plane so that the body remains at rest with respect to it.

### Step 1: Understand the Setup
We have an inclined plane making an angle θ with the horizontal. The height (h) of the incline is given as 1 unit, and the base (b) can be derived from the geometry of the triangle formed by the incline.

### Step 2: Identify Forces Acting on the Body
The forces acting on the body are:
1. The gravitational force (mg) acting vertically downward.
2. The normal force (N) acting perpendicular to the inclined plane.
3. A pseudo force (ma) acting horizontally in the opposite direction of the acceleration of the inclined plane.

### Step 3: Resolve the Gravitational Force
The gravitational force can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

### Step 4: Apply Newton's Second Law
For the body to remain at rest with respect to the inclined plane, the net force acting along the incline must be zero. Thus, we can write:
\[ mg \sin \theta = ma \cos \theta \]
Where:
- \( m \) is the mass of the body,
- \( a \) is the horizontal acceleration of the inclined plane.

### Step 5: Simplify the Equation
We can cancel the mass \( m \) from both sides:
\[ g \sin \theta = a \cos \theta \]
From this, we can express the acceleration \( a \):
\[ a = g \frac{\sin \theta}{\cos \theta} = g \tan \theta \]

### Step 6: Determine \( \tan \theta \)
From the geometry of the inclined plane:
- The height \( h = 1 \)
- The base \( b \) can be calculated using the Pythagorean theorem. If the hypotenuse is \( x \), then:
\[ b = \sqrt{x^2 - 1} \]
Thus:
\[ \tan \theta = \frac{h}{b} = \frac{1}{\sqrt{x^2 - 1}} \]

### Step 7: Substitute \( \tan \theta \) Back into the Acceleration Equation
Now substituting \( \tan \theta \) back into the equation for \( a \):
\[ a = g \tan \theta = g \cdot \frac{1}{\sqrt{x^2 - 1}} \]

### Step 8: Final Result
Thus, the required horizontal acceleration \( a \) of the inclined plane is:
\[ a = \frac{g}{\sqrt{x^2 - 1}} \]